This is an integrable system: denoting $y=x'$ we get

\begin{equation}

H(x,y) := \frac{1}{2}y^2 \underbrace{-\frac{1}{5}x^5+\frac{5}{3}x^3 -4x}_{V(x)}=E

\label{A}

\end{equation}

wehre $V(x)=-\int f(x)\,dx$, $f(x)=x^4-5x^2+4$ is the r.h.e. $V(x)$ is a potential.

Since $V(x)$ has non-degenerate maxima at $x=-1$ and $x=2$ and minima at $x=-2$ and $x=1$ and $\frac{1}{2}y^2$ has non-degenerate minimum at $y=0$ we have 4 non-degenerate critical points of $H(x,y)$: namely, $(-1,0)$ and $(2,0)$ are saddle points and $(-2,0)$ and $(1,0)$ are minima.

For dynamics two former are saddle points and two latter re centers (recall that **integrable systems cannot have spiral or nodal points and centers are detectable!**)

What is missing for everyone? Plot of $V(x)$ which shows that $ V(-1)>V(-2)>V(2)>V(1)$ (one needs just calculates them) and therefore picture must be like on Jason' computer generated and not Yeong' drawing since separatrix passing through saddle $(-1,0)$ is "higher" and therefore envelops saddle $(2,0)$ and cannot pass through it as Yeong drew. And therefore separatrix passing through $(2,0)$ stops short and goes back to the right from $(-1,0)$ (not passing through it). If in some examples we studied before separatrices passing through different saddles are the same it is not the general rule.

To analyze Jason's picture:

a) Look at the centers

b) Find separatrix passing through $(2,0)$

c) Find separatrix passing through $(-1,0)$